\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\) [568]

Optimal result

Integrand size = 33, antiderivative size = 137 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 b \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d} \]

[Out]

1/2*(2*A*b^2+a^2*(A+2*C))*arctanh(sin(d*x+c))/a^3/d-2*b*(A*b^2+C*a^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a
+b)^(1/2))/a^3/d/(a-b)^(1/2)/(a+b)^(1/2)-A*b*tan(d*x+c)/a^2/d+1/2*A*sec(d*x+c)*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3135, 3134, 3080, 3855, 2738, 211} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {A b \tan (c+d x)}{a^2 d}-\frac {2 b \left (a^2 C+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (a^2 (A+2 C)+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 a d} \]

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x]),x]

[Out]

(-2*b*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*Sqrt[a - b]*Sqrt[a + b]*d) + ((
2*A*b^2 + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (A*b*Tan[c + d*x])/(a^2*d) + (A*Sec[c + d*x]*Tan[c
+ d*x])/(2*a*d)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3080

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {A \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int \frac {\left (-2 A b+a (A+2 C) \cos (c+d x)+A b \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a} \\ & = -\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int \frac {\left (2 A b^2+a^2 (A+2 C)+a A b \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2} \\ & = -\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (b \left (A b^2+a^2 C\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^3}+\frac {\left (2 A b^2+a^2 (A+2 C)\right ) \int \sec (c+d x) \, dx}{2 a^3} \\ & = \frac {\left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (2 b \left (A b^2+a^2 C\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^3 d} \\ & = -\frac {2 b \left (A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (2 A b^2+a^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {A b \tan (c+d x)}{a^2 d}+\frac {A \sec (c+d x) \tan (c+d x)}{2 a d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.05 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (-2 \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \left (2 A b^2+a^2 (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 b \left (A b^2+a^2 C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (b \sin (c)+(-a+b \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right ) (i \cos (c)+\sin (c))}{\sqrt {\left (-a^2+b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 a A b \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {a^2 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 a A b \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{2 a^3 d (2 A+C+C \cos (2 (c+d x)))} \]

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + b*Cos[c + d*x]),x]

[Out]

(Cos[c + d*x]^2*(C + A*Sec[c + d*x]^2)*(-2*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ 2*(2*A*b^2 + a^2*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (8*b*(A*b^2 + a^2*C)*ArcTan[((I*Cos[c
] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(I*Cos[c] +
Sin[c]))/Sqrt[(-a^2 + b^2)*(Cos[c] - I*Sin[c])^2] + (a^2*A)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - (4*a*A*
b*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - (a^2*A)/(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^2 - (4*a*A*b*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(2*
a^3*d*(2*A + C + C*Cos[2*(c + d*x)]))

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-2 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(218\)
default	\(\frac {-\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (A \,a^{2}+2 A \,b^{2}+2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {A}{2 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (-A \,a^{2}-2 A \,b^{2}-2 a^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3}}+\frac {A \left (a +2 b \right )}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 b \left (A \,b^{2}+a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(218\)
risch	\(-\frac {i A \left ({\mathrm e}^{3 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d a}\)	\(498\)

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/2*A/a/(tan(1/2*d*x+1/2*c)+1)^2+1/2*(A*a^2+2*A*b^2+2*C*a^2)/a^3*ln(tan(1/2*d*x+1/2*c)+1)+1/2*A*(a+2*b)/
a^2/(tan(1/2*d*x+1/2*c)+1)+1/2*A/a/(tan(1/2*d*x+1/2*c)-1)^2+1/2/a^3*(-A*a^2-2*A*b^2-2*C*a^2)*ln(tan(1/2*d*x+1/
2*c)-1)+1/2*A*(a+2*b)/a^2/(tan(1/2*d*x+1/2*c)-1)-2*b*(A*b^2+C*a^2)/a^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/
2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 1.74 (sec) , antiderivative size = 535, normalized size of antiderivative = 3.91 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, {\left (C a^{2} b + A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} - 2 \, {\left (A a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, {\left (C a^{2} b + A b^{3}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{4} - A a^{2} b^{2} - 2 \, {\left (A a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[-1/4*(2*(C*a^2*b + A*b^3)*sqrt(-a^2 + b^2)*cos(d*x + c)^2*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c
)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x
+ c) + a^2)) - ((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + ((A + 2*C)
*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 - 2*(A*a^3*b
- A*a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c)^2), -1/4*(4*(C*a^2*b + A*b^3)*sqrt(a^2
- b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 - ((A + 2*C)*a^4 + (A - 2*C
)*a^2*b^2 - 2*A*b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + ((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*cos(
d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(A*a^4 - A*a^2*b^2 - 2*(A*a^3*b - A*a*b^3)*cos(d*x + c))*sin(d*x + c))/(
(a^5 - a^3*b^2)*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+b*cos(d*x+c)),x)

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**3/(a + b*cos(c + d*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.77 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {4 \, {\left (C a^{2} b + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

1/2*((A*a^2 + 2*C*a^2 + 2*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (A*a^2 + 2*C*a^2 + 2*A*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1))/a^3 + 4*(C*a^2*b + A*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arcta
n(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^3) + 2*(A*a*tan(1/2*
d*x + 1/2*c)^3 + 2*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1/2*c) - 2*A*b*tan(1/2*d*x + 1/2*c))/((tan(1
/2*d*x + 1/2*c)^2 - 1)^2*a^2))/d

Mupad [B] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 3926, normalized size of antiderivative = 28.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b*cos(c + d*x))),x)

[Out]

(A*a*sin(c + d*x))/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b*sin(2*c + 2*d*x))/(2*d*(a^2 - b^2)*(cos
(2*c + 2*d*x)/2 + 1/2)) - (A*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*d*(a^2 - b^2)*(cos(2*c
+ 2*d*x)/2 + 1/2)) - (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(d*(a^2 - b^2)*(cos(2*c + 2*d*x
)/2 + 1/2)) - (A*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)
/2 + 1/2)) + (A*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/
2 + 1/2)) + (C*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 +
1/2)) + (A*b^3*sin(2*c + 2*d*x))/(2*a^2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*a*atan((sin(c/2 + (d*x
)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(2*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (C*a*atan((si
n(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*
b^2*sin(c + d*x))/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2
- a^2)^(1/2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*A^2*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2
) + 4*C^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*A*C*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*
b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*a^2*b^7*sin(
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 3*A^2*a^5*b^4*sin(c/2
+ (d*x)/2)*(b^2 - a^2)^(1/2) - 2*A^2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A^2*a^7*b^2*sin(c/2 + (d
*x)/2)*(b^2 - a^2)^(1/2) + 8*C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*C^2*a^4*b^5*sin(c/2 + (d*x)/
2)*(b^2 - a^2)^(1/2) + 12*C^2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*
(b^2 - a^2)^(1/2) - 4*A*C*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2
- a^2)^(3/2) - 16*A*C*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 -
a^2)^(1/2) - 4*A*C*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^
7 + 4*C^2*a^7 - 3*A^2*a^3*b^4 + 2*A^2*a^5*b^2 - 4*C^2*a^5*b^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*(-(a + b)*(a - b)
)^(1/2)*1i)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*b^3*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2
)^(1/2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*A^2*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*
C^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*A*C*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*b*sin(
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*a^2*b^7*sin(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 3*A^2*a^5*b^4*sin(c/2 + (d*x
)/2)*(b^2 - a^2)^(1/2) - 2*A^2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A^2*a^7*b^2*sin(c/2 + (d*x)/2)
*(b^2 - a^2)^(1/2) + 8*C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 8*C^2*a^4*b^5*sin(c/2 + (d*x)/2)*(b^
2 - a^2)^(1/2) + 12*C^2*a^6*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 -
a^2)^(1/2) - 4*A*C*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)
^(3/2) - 16*A*C*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(
1/2) - 4*A*C*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^7 + 4*
C^2*a^7 - 3*A^2*a^3*b^4 + 2*A^2*a^5*b^2 - 4*C^2*a^5*b^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*(-(a + b)*(a - b))^(1/2
)*1i)/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) - (A*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))
*cos(2*c + 2*d*x)*1i)/(2*a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (A*b^4*atan((sin(c/2 + (d*x)/2)*1i)/cos
(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b^2*atan((sin(c/2 +
(d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x)*1i)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) + (C*b*cos(
2*c + 2*d*x)*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3
/2) - 8*A^2*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*C^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*A*C*
a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^8*b*sin(c/
2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^5*sin(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2) - 3*A^2*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*A^2*a^6*b^3*sin(c/2 + (d*x
)/2)*(b^2 - a^2)^(1/2) + 2*A^2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*C^2*a^4*b^3*sin(c/2 + (d*x)/2)
*(b^2 - a^2)^(3/2) - 8*C^2*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 12*C^2*a^6*b^3*sin(c/2 + (d*x)/2)*(b
^2 - a^2)^(1/2) - 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*A*C*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a
^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 16*A*C*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2
)^(1/2) + 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*A*C*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(
1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^7 + 4*C^2*a^7 - 3*A^2*a^3*b^4 + 2*A^2*a^5*b^2 - 4*C^2*a^5*b
^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*(-(a + b)*(a - b))^(1/2)*1i)/(a*d*(a^2 - b^2)*(cos(2*c + 2*d*x)/2 + 1/2)) +
(A*b^3*cos(2*c + 2*d*x)*atan(((A^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*b^7*sin(c/2 + (d*x)/2)*(b^
2 - a^2)^(3/2) - 8*A^2*b^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 4*C^2*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/
2) + 4*A*C*a^9*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - A^2*a^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a
^8*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 3*A^2*a^4*b^5
*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 3*A^2*a^5*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 2*A^2*a^6*b^3*sin
(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 2*A^2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 8*C^2*a^4*b^3*sin(c/2
+ (d*x)/2)*(b^2 - a^2)^(3/2) - 8*C^2*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) + 12*C^2*a^6*b^3*sin(c/2 +
(d*x)/2)*(b^2 - a^2)^(1/2) - 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*A*C*a^8*b*sin(c/2 + (d*x)/
2)*(b^2 - a^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2) - 16*A*C*a^2*b^7*sin(c/2 + (d*x)/2)
*(b^2 - a^2)^(1/2) + 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2) - 4*A*C*a^5*b^4*sin(c/2 + (d*x)/2)*(b
^2 - a^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^7 + 4*C^2*a^7 - 3*A^2*a^3*b^4 + 2*A^2*a^5*b^2 -
4*C^2*a^5*b^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*(-(a + b)*(a - b))^(1/2)*1i)/(a^3*d*(a^2 - b^2)*(cos(2*c + 2*d*x)
/2 + 1/2))